MongoDB query multiple collections at once

Question

users
{
 "_id":"12345",
 "admin":1
},
{
 "_id":"123456789",
 "admin":0
}

posts
{
 "content":"Some content",
 "owner_id":"12345",
 "via":"facebook"
},
{
 "content":"Some other content",
 "owner_id":"123456789",
 "via":"facebook"
}

Here is a sample from my mongodb. I want to get all the posts which has "via" attribute equal to "facebook" and posted by an admin ("admin":1). I couldn't figure out how to acquire this query. Since mongodb is not a relational database, I couldn't do a join operation. What could be the solution ?

Raul

Perform an Uncorrelated Subquery with $lookup mongodb.com/docs/manual/reference/operator/aggregation/lookup/… – Investiture

Answer 1

You can use $lookup ( multiple ) to get the records from multiple collections:

Example:

If you have more collections ( I have 3 collections for demo here, you can have more than 3 ). and I want to get the data from 3 collections in single object:

The collection are as:

db.doc1.find().pretty();

{
    "_id" : ObjectId("5901a4c63541b7d5d3293766"),
    "firstName" : "shubham",
    "lastName" : "verma"
}

db.doc2.find().pretty();

{
    "_id" : ObjectId("5901a5f83541b7d5d3293768"),
    "userId" : ObjectId("5901a4c63541b7d5d3293766"),
    "address" : "Gurgaon",
    "mob" : "9876543211"
}

db.doc3.find().pretty();

{
    "_id" : ObjectId("5901b0f6d318b072ceea44fb"),
    "userId" : ObjectId("5901a4c63541b7d5d3293766"),
    "fbURLs" : "http://www.facebook.com",
    "twitterURLs" : "http://www.twitter.com"
}

Now your query will be as below:

db.doc1.aggregate([
    { $match: { _id: ObjectId("5901a4c63541b7d5d3293766") } },
    {
        $lookup:
        {
            from: "doc2",
            localField: "_id",
            foreignField: "userId",
            as: "address"
        }
    },
    {
        $unwind: "$address"
    },
    {
        $project: {
            __v: 0,
            "address.__v": 0,
            "address._id": 0,
            "address.userId": 0,
            "address.mob": 0
        }
    },
    {
        $lookup:
        {
            from: "doc3",
            localField: "_id",
            foreignField: "userId",
            as: "social"
        }
    },
    {
        $unwind: "$social"
    },

  {   
    $project: {      
           __v: 0,      
           "social.__v": 0,      
           "social._id": 0,      
           "social.userId": 0
       }
 }

]).pretty();

Then Your result will be:

{
    "_id" : ObjectId("5901a4c63541b7d5d3293766"),
    "firstName" : "shubham",
    "lastName" : "verma",

    "address" : {
        "address" : "Gurgaon"
    },
    "social" : {
        "fbURLs" : "http://www.facebook.com",
        "twitterURLs" : "http://www.twitter.com"
    }
}

If you want all records from each collections then you should remove below line from query:

{
            $project: {
                __v: 0,
                "address.__v": 0,
                "address._id": 0,
                "address.userId": 0,
                "address.mob": 0
            }
        }

{   
        $project: {      
               "social.__v": 0,      
               "social._id": 0,      
               "social.userId": 0
           }
     }

After removing above code you will get total record as:

{
    "_id" : ObjectId("5901a4c63541b7d5d3293766"),
    "firstName" : "shubham",
    "lastName" : "verma",
    "address" : {
        "_id" : ObjectId("5901a5f83541b7d5d3293768"),
        "userId" : ObjectId("5901a4c63541b7d5d3293766"),
        "address" : "Gurgaon",
        "mob" : "9876543211"
    },
    "social" : {
        "_id" : ObjectId("5901b0f6d318b072ceea44fb"),
        "userId" : ObjectId("5901a4c63541b7d5d3293766"),
        "fbURLs" : "http://www.facebook.com",
        "twitterURLs" : "http://www.twitter.com"
    }
}

Glowing

Hi, @Shubham Verma, how to get the multiple records from "social" or "address" collections. in the same query. like I have multiple records in social collections for the single user – Siloum

It is a very nice answer. Simple, clear and limpid! – Kirovograd

Answer 2

Trying to JOIN in MongoDB would defeat the purpose of using MongoDB. You could, however, use a DBref and write your application-level code (or library) so that it automatically fetches these references for you.

Or you could alter your schema and use embedded documents.

Your final choice is to leave things exactly the way they are now and do two queries.

Puleo

How would one go about writing two queries for this? – Exuberance

Doctrine MongoDB ODM is a decent library for handling such things. However it only lets you join through one layer of references; you can't populate references within references within references with any degree of syntactical ease. – International

How could this be achieved with plain old queries in MongoJS? – Variole

You can now use $lookup to join – Moderato

Sorry, but I just have to say, "Trying to use a DB without a concept of a JOIN defeats the purpose of databases. Entirely" Mongo added $lookup due to mistakenly assuming this not to be the case. Embedded documents are not the best answer as they are not synchronized with the records they should be equivalent to, and you wind up duplicating data. – Coughlin

An important point to highlight here is that DBRefs are implemented by client side drivers which implies multiple fetch calls. As suggested above, use $lookup to avoid multiple network trips. – Bandsman

How can i delete user 12345 from users collection and all posts created by the user at once. – Adlai

Answer 3

Here is answer for your question.

db.getCollection('users').aggregate([
    {$match : {admin : 1}},
    {$lookup: {from: "posts",localField: "_id",foreignField: "owner_id",as: "posts"}},
    {$project : {
            posts : { $filter : {input : "$posts"  , as : "post", cond : { $eq : ['$$post.via' , 'facebook'] } } },
            admin : 1

        }}

])

Or either you can go with mongodb group option.

db.getCollection('users').aggregate([
    {$match : {admin : 1}},
    {$lookup: {from: "posts",localField: "_id",foreignField: "owner_id",as: "posts"}},
    {$unwind : "$posts"},
    {$match : {"posts.via":"facebook"}},
    { $group : {
            _id : "$_id",
            posts : {$push : "$posts"}
    }}
])

Lovelady

What this does? Is this a query or does it add that metadata to the collection? – Finespun

The $lookup operator is new since version 3.2 and is like a left outer join. See docs.mongodb.com/manual/reference/operator/aggregation/lookup – Visser

Hi, can you please let me know what does filter do here in the first query? – Patinous

Answer 4

As mentioned before in MongoDB you can't JOIN between collections.

For your example a solution could be:

var myCursor = db.users.find({admin:1});
var user_id = myCursor.hasNext() ? myCursor.next() : null;
db.posts.find({owner_id : user_id._id});

See the reference manual - cursors section: http://es.docs.mongodb.org/manual/core/cursors/

Other solution would be to embed users in posts collection, but I think for most web applications users collection need to be independent for security reasons. Users collection might have Roles, permissons, etc.

posts
{
 "content":"Some content",
 "user":{"_id":"12345", "admin":1},
 "via":"facebook"
},
{
 "content":"Some other content",
 "user":{"_id":"123456789", "admin":0},
 "via":"facebook"
}

and then:

db.posts.find({user.admin: 1 });

Visser

Answer 5

Perform multiple queries or use embedded documents or look at "database references".

Biebel

Answer 6

One solution: add isAdmin: 0/1 flag to your post collection document.

Other solution: use DBrefs

Donnetta

Answer 7

Posting since I wanted to flatten the merged documents, vs a tiered document that the other answers produce.

To merge multiple collections into a flat single document, look at Mongo docs for $lookup with $mergeObjects: https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/#use--lookup-with--mergeobjects

Aciculum